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I'm trying to solve the following problem:

What is the number of different triangles we can form from numbers $4,5,6,7,8,9$ (lengths of sides), where every side has a different length (for an example $4,5,6$ or $4,5,7$)..

My solution is the following:

The number of all possible permutations is $\frac{6!}{(6-3)!}$. We subtract the numbers which don't make a triangle (which is $6*2$) and then divide by $2$, because $(4,5,6)$ makes the same triangle is $6,5,4$. My answer is $54$.

However, the correct answer should be $53$. Can anyone tell me where I did a mistake?

Thanks

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    $\begingroup$ $6 \choose 3$ is only $20$ and there is one choice $(4,5,9)$ that doesn't make a triangle, so I get $19$ Your expression (ignoring the factorial in the number to choose) gives $8$. $\endgroup$ Oct 20, 2019 at 16:42
  • $\begingroup$ I think you'll have to show more details for someone to point out your mistake. What do you mean by $\binom{6}{(6-3)!} $? It look to me like ${6\choose6}=1$. Did you mean to say ${6\choose3}?$ Also, why do you say there are $12$ where the triangle inequality doesn't hold. How did you get that? $\endgroup$
    – saulspatz
    Oct 20, 2019 at 16:43
  • $\begingroup$ Sorry, I made a mistake when writing my answer here. Please wait a minute and I will fix it $\endgroup$
    – Peter F.
    Oct 20, 2019 at 16:44
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    $\begingroup$ If you write out all the triples $(4,5,6), (4,5,7), \ldots, (7,8,9)$, you'll see that there are only $20$ in the list, one of which is degenerate. How can there be $53$ or $54$ triangles? We could consider "right-handed" and "left-handed" triangles to be different, effectively doubling the number of triangles, but even then, we would have at most $40$, still falling short of the given answer. Have you written out the problem exactly as it was given to you? $\endgroup$
    – Théophile
    Oct 20, 2019 at 16:56
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    $\begingroup$ You should have gotten six crossed off, as you describe. 2 for 4, 2 for 5 and 2 for 9 (not 6) makes six. Why do you double that? Then you should divide by $6$ because 456 comes six ways-456,465,546,564,654,645 not two $\endgroup$ Oct 20, 2019 at 17:04

1 Answer 1

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There are $\frac {6!}{3!}$ ordered choices of three numbers, which is $120$. Six of those do not make a triangle, all the permutations of $(4,5,9)$, which leaves $114$. Each unordered triangle gives $3!$ permutations, so we divide by $6$ and get $19$. I don't know where a number in the $50$s comes from.

I think it is easier to just choose unordered combinations to start with, which is ${6 \choose 3}=20$ and subtract the one that doesn't make a triangle. That also gets $19$.

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